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Perpendicular/Parallel Lines

Date: 01/17/97 at 16:59:36
From: John Hartge
Subject: Slope-intercept forms

My daughter is having trouble with the slope intercept form of lines.  
Given a line in this form, she has trouble finding a line perpendicuar 
or parallel to it. She also doesn't understand how to find the x and y 
intercepts from an equation like 4x + 3y = 6 or its graph.  When she 
was doing mixture problems earlier this year, she kept having trouble 
writing a table to figure it out.  Can you please help me?

Thanks, John

Date: 01/21/97 at 22:46:33
From: Doctor Keith
Subject: Re: Slope-intercept forms


That is a pretty good list of things; glad to see a dad helping his 
kid! Hopefully I will be able to clear this up.  I have written it so 
your daughter can follow it (it is a bit long since I am quite wordy, 
so you probably will want to print it out).  

I have tried to guess at exactly what she knows and does not know, so 
please reply if something needs clarification and forgive things that 
might be too simple.  

Now we are lucky in that I think we can handle most of the questions 
by doing a good review of equations of a line.

The slope intercept form of a line is:

    y = mx + b

  y is the vertical measurement (sometimes called dependant variable)
  x is the horizontal measurement (sometimes called independant    
  m is the slope (the change in the y divided by the change in x which 
    is sometimes expressed as rise over run)
  b is the y-intercept

I advise you to put any linear equation into this format so that you 
can then use some standard rules and tools to get your answers.  Thus 
if you had a problem like:

      3y = 5/3x - 8

you would rewrite it to read (by dividing by 3):

       y = 5/9x - 8/3

Now let's tackle the easiest part of the problem, which is finding the
x and y intercepts.  My math teacher taught us what he called the 
Watergate method (for obvious reasons as you will see).  The method
goes like this:

To find the y-intercept:

  1) Take your hand (or finger or whatever) and cover-up the x term
     (that is mx). (The cover-up is the reason for the name 
  2) The solution is y = b (x = 0).
Note this works because with x = 0, we have mx = 0 and thus y = b, and
since the y intercept is the value of y when x = 0, we are done.

To find the x-intercept:

  1) Take your hand (or finger or whatever) and cover-up the y term 
     (that is just y this time).
  2) We have 0 = mx + b.
  3) Thus solving for x we have x = -b/m.
Note this works because with y = 0, we have step (2), and the 
x-intercept is the value of x when y = 0 so again we are done.

Thus in the problem above we have:

               y = 5/9x - 8/3
     y-intercept = b = -8/3
     x-intercept = -b/m = (8/3)(9/5) = 24/5

Now the next easiest thing is getting a parallel line.  Actually, 
since we know how to get the intercepts, we can get the parallel lines 
just as easily.  Note that all parallel lines have the same slope.  
This makes sense because both lines must increase the same amount in y 
for a given change in x or at some point they would meet (this is easy 
to convince yourself of by drawing some lines on a piece paper).  

Now if you draw a bunch of parallel lines, you will notice the only 
difference between them is where they intercept the x or y axis. So by 
changing, say, the y-intercept, you can create a new parallel line.  
We know that the y-intercept is b in our equation so to get a parallel 
line you just need to change b. 

Example of line parallel to y = 5/9x - 8/3:

    y = 5/9x - 1

Now say you wanted a line parallel through a point (a bit trickier but 
still no problem).  You are given (x1,y1) and you need to find a b1 
for which y1 = mx1 + b1 is valid.  Thus you can solve for b1 and you 

     b1 = y1 - m(x1)

Example line parallel to y = 5/9x - 8/3 through (9,1):

     b = 1 -5/9(9) = -4
     y = 5/9x -4

Okay, now you need to know how to find perpendicular lines also.  
Even more fun.  This relies on the neat property of slopes that the 
product of the slopes of two perpendicular lines is -1.  The easiest 
way I know to prove this involves some higher math that you don't want 
to deal with at the moment (no reason to muddy the waters) but you can 
see that this makes sense by the following:

  1) Draw two perpendicular lines on a blank sheet of paper (no axis 
     please - we are going to rotate the paper to compare slopes).
  2) We shall agree that all lines are drawn left to right regardless 
     of the up/down motion.
  3) Thus we have that lines are drawn UP if they start in the lower 
     left and go to the upper right. Note these lines will have 
     positive slopes (the y value increases as x increases).
  4) Similarly, lines that are drawn DOWN (upper left to lower right)
     have negative slopes (the y values decrease if x increases).
  5) Note that a negative times a positive must be a negative, so we 
     know from (3) and (4) that the products of the slopes must be 

  6) Rotate the paper in front of you. You will notice that as one 
     line gets steeper (large change in y for a given change in x), 
     the other line gets shallower (small change in y for a given 
     change in x).  This indicates that the slopes are inversely 
     proportional. In other words, the product of the slopes of the 
     lines should be a constant regardless of how you rotate the 
  7) So far, we have that the product of the slopes should be a 
     negativeconstant.  To see why the constant is -1, do the 
     following. Turn the paper so that the lines are at a 45 degree 
     angle with the horizontal (it should look like an x). Now the 
     slopes are 1 and -1 for this case. We have noted that the slopes 
     are a constant regardless of the rotation [i.e., the product of 
     the slopes gives the same answer for our 45 degree case as for 
     any other rotation angle (except for a constant line y = 0 and 
     x = 0, since x = 0 is not allowed in math)]. Since the product in 
     this case is -1, it must hold for the other cases.  As I 
     indicated earlier, while this does not prove it (the proof is 
     beyond the scope of this discussion) it helps you to see that it 
     is true.

Now, using this you can get a perpendicular line by the following:

     let your first line be y = m1x + b1
     you want a line        y = m2x + b2  that is perpendicular

Solve by using the fact that the product of the slopes is -1:

     m1(m2) = -1 
     m2 = -1/m1

Then if you want the line to go through a particular point (x1,y1), 
you can find b2 by (similar to the parallel case):

     y1 = m2x1 + b2
     b2 = y1 - m2x1 
        = y1 + x1/m1

If you do not have to go through a particular point, pick any number 
for b2 (I suggest b2 = 0).

As for general graphing and writing the tables to graph form, I 
suggest the following:

  1) Pick equally spaced numbers usually centered on the origin to use 
     as your x values.  Try to pick them so your calculations are easy 
     (i.e., for a slope of 1/3 values of -3,0,3,6,... would be great).

  2) Make a table that looks like   x   |   y     and put the x values
                                  --------------  in the left column
  3) For each value in the x column, calculate y = mx + b for y (i.e., 
     you know m, x, and b) and put this in the y column (this can be
     simplified in linear equations by noting that by multiplying the 
     change in x times the slope and adding it to the previous value 
     you will get the new value, see example below).

  4) When you are done with the table, just plot the points and 
     connect the dots.

Example: y = 2/3 x - 4

  1) I pick -3,0,3,6,9

  2)        x   |    y
           -3   |
            0   |
            3   |
            6   |
            9   |
  3)        x   |    y
           -3   |    2/3(-3) -4 = -6
            0   |    2/3(0) - 4 = -4      (or -6 + 2/3(0--3) = -6 +2 )
            3   |    2/3(3) - 4 = -2      (or -4 + 2/3(3-0) = -4 +2 )
            6   |    2/3(6) - 4 = 0       (or -2 + 2 )
            9   |    2/3(9) - 4 = 2       (or 0 + 2 )

  So you have:
            x   |    y
           -3   |    -6
            0   |    -4 
            3   |    -2 
            6   |    0
            9   |    2

Hope this helps.  Keep up the good work and let me know how it all 
works out.  We are here to help.  Good luck!

-Doctor Keith,  The Math Forum
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Associated Topics:
Middle School Equations

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